Aptitude - Pipes and Cistern - Discussion

Discussion Forum : Pipes and Cistern - General Questions (Q.No. 11)
11.
One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
81 min.
108 min.
144 min.
192 min.
Answer: Option
Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in x minutes.
3

1 + 3 = 1
x x 36

4 = 1
x 36

x = 144 min.

Discussion:
67 comments Page 3 of 7.

Indrajit said:   1 decade ago
Slower pipe fills in x min ,faster fills x/3.
Let for complete the tank use symbol C.

For slower pipe:
in x min it fills C
in 1 min it fills C/x

Similarly in 1min faster pipe fills 3C/x
Both fills in 36min
so,in 36min both fills C
in 1 min both fills C/36

Now, C/x+3C/x=C/36
or 1/x+3/x=1/36

Hope you got it!

Karthik said:   1 decade ago
First pipe fills 3 times faster than second one so x=3x
If together the two pipes can fill the tank in 36 minutes so x*3x=36
x=36/3x
x=12/x
x=144

Kant said:   1 decade ago
Let the volume b unit meter cube
so rate of individual filling = rate of combine fillin
so
1/X +1/(X/3) = 1/36
so x=144

Ali said:   1 decade ago
@Karthik how it is x square=12 and x =144???

Yasin said:   1 decade ago
I AM TRYING LIKE THIS but not getting.

First pipe x to fill tank then second one is 3x.
now both 1/x+1/3x=36
4/3x=36
1/x=9???
What's the problem?

Mahima said:   1 decade ago
To all those who have not understood yet.

Try this.

It is given that- One pipe can fill a tank three times as fast as another pipe.

Not let 1st pipe is A (faster) and 2nd pipe is B (slower).

Now let pipe A fills the tank in x/3 minute.

Therefore in 1 minute pipe A will fill 1/ (x/3) part of the tank.

Similarly let pipe B fills the tank in x minute (because pipe B is slower therefore it will take more time).

Therefore in 1 minute pipe B will fill 1/x part of the tank.

Therefore in 1 minute both the pipe (A and B) fill 1/36 part of the tank.

Now it is given that- Together the two pipes can fill the tank in 36 minutes.

Therefore A+B=36 mins.

Substitute the value of A and B according to 1 min we get.

=> 1/ (x/3) + 1/x =1/36.

=> 3/x +1/x= 1/36.

=> 4/x =1/36.

=> 4*36=x.

=>x=144 (ans).

Manoj said:   1 decade ago
So simple,

Time=work/capacity,

So total capacity=1/x+3/x.

And total work=1.

So total time taken by both of them=36=1/(1/x+3/x).

Harshdeep said:   1 decade ago
Another simpler approach:

(slower) pipe A takes 'x' min to fill complete tank alone.
Then pipe A in 1 min fills (1/x) portion of tank alone.

(faster) pipe B takes (x/3) min to fill the complete tank alone.
then pipe B in 1 min fills (3/x) portion of tank alone.

When both pipes together fills the tank then in 1 min they fill (1/x + 3/x) portion of tank.

This means complete tank can be filled in (x + x/3) minutes.
(4x/3) = 36.... so x = 144 min . Answer.

Ravikanth said:   1 decade ago
The best approach to these kinda questions are the logic of replacing the variables with the numbers but be careful in doing so because one should be aware of the numbers which are being replaced to the variables should be satisfying the condition of the variables.

So guys the easiest approach to the above problem is to:
P1 is three times faster than P2 right?

So I will write down a statement as follows,
P1=3P2.

So am gonna replace these with numbers as below,
P1=3. If p1=3 the above statement becomes 3=3P2.

By the cancellation we get the value of P2 as 1,
P2=1.

If a tank is to be filled by the two pipes together in 36 minutes then the values P1=3 becomes 3L/M P2 = 1L/M.

So my next statement is,
P1+P2 = 3+1.
P1+P2 = 4L/M.

If both the pipes are working together then the tank is being filled at a rate of 4L/M.

So the time taken by the to pipes to fill the tank is 36M.

Then we need to multiply the combined rate of pipes with the time to get the capacity of the tank.

4L/M*36M=144L which the capacity of the tank.

Now coming to the last part of the question.

"slower pipe alone will be able to fill the tank in:"

I already mentioned the slower pipe as P2 and the rate of P2=1L/M so the time taken by P2 to fill the tank of 144L capacity is (tank capacity/P2)=144L/1L/M=144M.

And finally sorry for the big explanation and guys please pardon me if there are any grammatical errors.
In brief,

P1 = faster pipe P2 = slower pipe.
P1=3P2.

If P1 = 3 =>3 = 3P2 =>P2 = 1.
Time = 36M.

P1+P2 = 1+3 =>P1+P2 = 4.
As the units of the pipes are liters/minute L/M
Find tank capacity,

(P1+P2)*Time=4*36=>Tank capacity = 144L.

144L capacity tank will be filled by slower pipe at its slower rate right?
rate of slower pipe P2?

P2 = 1L/M.

Time=Tank capacity/rate of pipe P2 => Time = 144/1 = 144.

Which is the required answer.

Anil kumar said:   1 decade ago
If slower pipe is x. Than faster 3x. Take both 36 min. means x = 9 let 9+3 = 12 and 12 square = 144.


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